Tài liệu Physics exercises_solution: Chapter 25 ppt

25.29: a If 120 strands of wire are placed side by side, we are effectively increasing the area of the current carrier by 120.. 25.33: a An ideal voltmeter has infinite resistance, so th

25.1: Q  It (3.6A)(3)(3600s)3.89104 C.

) 60 ( 80 C

C106.1)(

108.5(

A1075.8

2 3 19

4π)(

C106.1)(

105.8(

A85.4

2 3 19





b) If the diameter is now 4.12 mm, the time can be calculated using the formula above

or comparing the ratio of the areas, and yields a time of 26542 s =442 min

c) The drift velocity depends on the diameter of the wire as an inverse square

relationship

25.4: The cross-sectional area of the wire is

.m10333.1)m1006.2

πr π A

The current density is

2 5 2

5 6.00 10 A mm

10333.1

A00

Wehavev d J ne;Therefore

19 5

2 5

m

electrons10

94.6)electronC

1060.1)(

sm1040.5(

mA1000

sm1020.1()()

1 2 1 1 2

25.6: The atomic weight of copper is 63.55g mole,and its density is 8.96g cm3 The number of copper atoms in 1.00m3 is thus

moleg55.63

)moleatoms10

023.6)(

mcm1000.1)(

cmg96.8

3

28 atoms m10

49

Since there are the same number of free electrons m3 as there are atoms of copper m3(see Ex 25.1), The number of free electrons per copper atom is one

25.7: Consider 1 m of silver.3

density10.5103 kg m3,som10.5103 kg

M 107.86810 3kg mol,sonm M 9.734104 moland

3 28

C0106

65.055

0 3 8

0 8

0

2 8

1333

s100.8A

6.3

)m103.2)(

C106.1)(

m105.8)(

m0.4

4(

)m0.24)(

m10

72.1(

2 3

ρL

R

25.12: 9.75m.

m10

72.1

)m10462.0)(

4)(

00.1(

8

2 3

ρL

R

25.13: a) tungsten:

.mV1016.5)

m1026.3)(

4(

)A820.0)(

m10

25.5

2 3

3 8

ρI ρJ

E

b) aluminum:

.mV1070.2)

m1026.3)(

4(

)A820.0)(

m10

75.2

2 3

3 8

ρI ρJ

E

25.14:

Al

Cu Al Cu Cu

Cu Al

l A Cu

Cu Al

Al Cu

Al

ρ

ρ d d ρ

πd ρ

πd A

L ρ A

L ρ R

44

2 2

mm

6.2m10

75.2

m10

72.1)mm26.3

ρL

Ohm125.0

)m50.3)(

mOhm10

72.1

R

ρL AL

m(density)V (8.9103 kg m3)(1.68610 6 m3)15g

25.16:

mm625.12

25.17: a) From Example 25.1, an 18-gauge wire has A8.1710 3 cm2

A820)cm1017.8)(

A/cm10

0

100.1(

0 0

35.2

]C)20320)(

C105.4(1[

)]

(1[

ρ ρ

T T ρ

Since E J, the electric field required to maintain a given current density is

proportional to the resistivity Thus E(2.35)(0.0560V m)0.132V m

m80.1

m10

75.2

L

ρ L

ρL A

ρL

R

25.20: The ratio of the current at 20C to that at the higher temperature is

.909.3)A220

C666C105.4

1909.3C201

)(

1

3 0

0 0

T T ρ

ρ

25.21:

m

1005

2

)V50.1(

)m20.1)(

m10

75.2)(

A00.6(4

8 2

L I r πr

ρL A

ρL I

V

)m50.2)(

A6.17(

)m1054.6()V50.4

RA

ρ

25.23: a) 11.1A.

)m10

44.2(

))m1084.0(4)(

mV49.0(

8

2 3

)m1084.0)(

4(

)m4.6)(

m10

44.2)(

A1.11(

2 3

L I IR

A1.11

V13

2

)2(

0

R A

ρL A

L ρ

That is, four times the original resistance

m75.0

V938

RI L

RAJ ρJ

E

)m75.0)(

mA1040.4(

V938

C0.20C0.34(

484.1512.1)

(

1 3 0

R R

i f



25.27:a)   (  ) 100100 (0.0004C 1)(11.5C)99.54

f i

f i i

b)   (  ) 0.01600.0160(0.0005C 1)(25.8C)

f i

f i i

R R T

R R T T

C0005.0(

3.2178

.215

1

o o

25.29: a) If 120 strands of wire are placed side by side, we are effectively increasing the

area of the current carrier by 120 So the resistance is smaller by that factor:

.1067.412010

60

V6

Change in terminal voltage:

r I

rI

V T

V2.2

V2.2V4.10V6.12

V6.12

Solve for r: r  8460 

)m050.0(

)m10100)(

m1072.1

2

3 8

π A

L

R 

V4.27)219.0)(

A125

J/s3422(

25.32: a) V r εV ab 24.0V21.2V2.8Vr2.8V 4.00A0.700 b) V R 21.2VR21.2V 4.00A5.30

25.33: a) An ideal voltmeter has infinite resistance, so there would be NO current

through the 2.0resistor

b) V ab ε5.0V; since there is no current there is no voltage lost over the internal resistance

c) The voltmeter reading is therefore 5.0 V since with no current flowing, it measures the terminal voltage of the battery

25.34: a) A voltmeter placed over the battery terminals reads the emf: ε24.0V b) There is no current flowing, so V r 0.

c) The voltage reading over the switch is that over the battery: V s 24.0V

d) Having closed the switch:

.V9.22)28.0)(

A08.4(V0.24A

08.488

A08

V11.0V

11.0V97.2V08



I

V r

1

V97

2

R

25.36: a) The current is counterclockwise, because the 16 V battery determines the

direction of current flow Its magnitude is given by:

A

47.00.94.10.56.1

V0.8V0

25.37: a) Now the current flows clockwise since both batteries point in that direction:

A

41.19.01.4

5.01.6

V0.8V0

48.5)

A21.0)(

)4.10.96.1((

V0

b) The Nichrome wire does obey Ohm’s Law since it is a straight line

c) The resistance is the voltage divided by current which is 3.88

25.40: a) Thyrite resistor:

b) The Thyrite is non-Ohmic since the plot is curved

c) Calculating the resistance at each point by voltage divided by current:

V15

V

25.43: P  VI(650V)(0.80A)520W

25.44: W Pt IVt(0.13A)(9V)(1.5)(3600s)6318J

25.45: a) ( ) since

vol

2 2

2 2 2

L

A L A J AL

R A J AL

R I P p R I

R

W

0.2)0.9()A47.0(and

kg056.0vol

g0.56gJ

000

,

46

J10

c) To recharge the battery:

.h6.1)W450()Wh720()

 Wh P t

25.48: a) I ε (Rr)12V101.2APεI (12V)(1.2A)14.4W

This is less than the previous value of 24 W

b) The work dissipated in the battery is just: PI2r(1.2A)2(2.0)2.9W.This is less than 8 W, the amount found in Example (25.9)

c) The net power output of the battery is 14.4W2.9W11.5W This is less than

16 W, the amount found in Example (25.9)

25.49: a) I V R12V 62.0APεI (12V)(2.0A)24W

b) The power dissipated in the battery is P  r I2 (2.0A)2(1.0)4.0W

c) The power delivered is then 24W4W20W

25.50: a) I  ε/R3.0V/170.18API2R0.529W

b) W Pt IVt (0.18A)(3.0V)(5.0)(3600s)9530J

c) Now if the power to the bulb is 0.27 W,

.8.6567

)17

()17(17

V0.3W27

I

P

25.51: a) PV2 RRV2 P(120V)2/540W26.7.

b) I V R120V/26.74.5A

c) If the voltage is just 110 V, then I 4.13APVI 454W

d) Greater The resistance will be less so the current drawn will increase, increasing the power

τ ne

m

.s1055.1m)2300(C)1060.1()m100.1(

kg1011

2 19 3

16

31 2

m τ

b) The number of free electrons in copper (8.51028 m 3) is much larger than in pure silicon (1.01016 m 3)

m0.14

m)1050.2()4()104.0

m10

65.3

m)1050.2()4()mV28.1(

8

2 3

I

c)

C)106.1()m10(8.5m)10

(3.65

V/m28.1

19 3

J

v d

m/s

1058

VA A ρl

V R

V

m)(25.0m)10

(1.47

m)10(0.100m)1000.2)(

2()V

12

(

8

3 2



25.55: With the voltmeter connected across the terminals of the battery there is no

current through the battery and the voltmeter reading is the battery emf; ε12.6 V

With a wire of resistance Rconnected to the battery current I flows and

I r I R

ε 12.6V(7.00A)r(7.00A)R10

;0)2( 12

V I

10(6.0/4)(

m)(0.8m)10

72.1(

2 4

L ρ R

Cu

Cu Cu Cu

m)(1.2m)10

47.1(

2 4

8

π A

L ρ R

Ag

Ag Ag Ag

A

45062

.0049.0

V0

So the current in the copper wire is 45 A

b) The current in the silver wire is 45 A, the same as that in the copper wire or else charge would build up at their interface

m8.0

)049.0()A45

Cu

L

IR Jρ

E

m2.1

)062.0()A45

Ag

L

IR Jρ

E

e) V Ag  IR Ag (45A)(0.062)2.79V

25.57: a) The current must be the same in both sections of the wire, so the current in the

thin end is 2.5 mA

)A106.1()4(

)A10(2.5m)10

72.1

2 3

3 8

ρI ρJ E

3 8

0.8mm

)A1080.0()4(

)A10(2.5m)10

72.1(

ρI ρJ E

=8.5510 5 V/m(4E1.6mm)

d) V E1.6mmL1.6mm E0.8mmL0.8mm

V.101.80m)(1.80V/m)10

(8.55m)

(1.20V/m)10

14.2

.m/J107

8

m/s)10(1.5kg)1011.9()m105.8(2

1volume

3 10

2 4 31

3 28

.106.1J107.8

J13600So

ρdx A

)

0

2 2 1 2

1 2

1

2

1 2

ρh R u

r r π ρh

u

du ) r π(r

ρh x

r π

dx ρ R

r

r

r h

r r

A

ρL πr

ρh R r r

4

144

)(

4)

a b ρ

πab V A

I J a b ρ

πab V R

,4

4

)(11

A

ρL πa

ρL πab

a b ρ b a π

Q

E ρJ

E

0 0

0 0

25.62: a)  /  V L.

A A

L V

RA V A I R

I         So to make the current density a

maximum, we need the length between faces to be as small as possible, which means

d

L So the potential difference should be applied to those faces which are a distance

dapart This maximum current density is ρd V

 must be a maximum The maximum area

is presented by the faces that are a distance dapart, and these two faces also have the greatest current density, so again, the potential should be placed over the faces a distance

dapart This maximum current is

.6

4(

m)(0.12m)10

5.9(

ρL R

109.83C)60

108.64C)

constant As the fluid expands the container, outward expansion “becomes” upward expansion due to surface effects

d)

A

L ρ A

ρL R A

ρL

R     

.1040.2

m)(0.0016/4)

(

m)10(0.86m)10

95(m)

(0.0016/4)

(

m)(0.12m)10

34.3(3

2

3 8

2 8

1 ( 0 057 0.0572 40 10 ) C

401

0

R R T

α

.)

25.64: a) 0.167A

0.24

V0.4V0

.24

V4.0V0.8V3

(0.50A)(0.257V

00

5.00

V8.4V9.4

A)(3.50)

A)(1.50V

(8.4V4.9

4.1A0.001

V000,

m)(0.10m)0.5(

ρL R

b) V IR(10010 3A)(1000)100V

c) PVI (100V)(10010 3A)10W

25.68: a) V 2.50I 0.360I2 4.0V Solving the quadratic equation yields

A,8.29orA34

(2.68A)/V50.2

6.12)2.38.3()3

1

(

0)

I

V I α R βI

4.285.0

V86.785

.0A25.9

V86

I I

R R r I R

R r

R I

R r

R R r I R r

A

A A

ε

b) We want:

.0425.0

)8.345.0()01.0(01

.001

.11

r

R R

r

R I

I

c) This is a maximum value, since any larger resistance makes the current even less that it would be without it That is, since the ammeter is in series, ANY resistance it has increases the circuit resistance and makes the reading less accurate

25.72: a) With a voltmeter in the circuit:

V R r

b) We want:

.6.44045

999901

.0

01.0

01.099

.01

R

R r

r R

r

r V

V

V V

ab

ε

c) This is the minimum resistance necessary—any greater resistance leads to less

current flow and hence less potential loss over the battery’s internal resistance

25.73: a) The line voltage, current to be drawn, and wire diameter are what must be

considered in household wiring

V120

carry up to 40 A

m)26003.0()4(

)m(42.0m)10

72.1()A35(

2

8 2

I

P

d) If 6-gauge wire is used,

.25.19

$)kWh11.0()kWh175(Savings

kWh175)h12()365()W40(

W66m)

(0.00412)

)4(

m)(42m)Ω10(1.72A)(35

2

8 2

π A

ρL I

P

25.74: Initially: R0 V I0 (120V) (1.35A)88.9

Finally: R f V I f (120V) (1.23A)97.6

.C237C20C217C

217

19.88

6.97C105.4

11

1)(

)(

0 0

f f

f f

T T

T

R

R α T T T

T α R

R

b) (i) P0  VI0 (120V)(1.35A)162W

(ii) P f VI f (120V)(1.23A)148W

25.75: a) 0.40A.

0.10

V8.0V0

(8.02

018.0()4(

m)(2.0m)10

0.2(

57.1()A15000()(

012.0)

m008.0()4(

)m(35m)10

72.1(

3 2

Cu

R R

I IR V

π A

q E q ma



b) If the electric field is constant, | |

bc bc

V

aL m

q EL

)C10(1.6V)100.1(

|

31

19 3

a bc

e) Performing the experiment in a rotational way enables one to keep the

experimental apparatus in a localized area—whereas an acceleration like that obtained in (d), if linear, would quickly have the apparatus moving at high speeds and large

distances

25.78: a) We need to heat the water in 6 minutes, so the heat and power required are:

W

233)

s60(6

J83800

J83800)

C80()CJ/kg4190()kg250.0(

T mc

W233

)V120

2 2

V P

m10

00.1

)m105.2()8.61(vol

3 5 2

L R

Now the radius of the wire can be calculated from the volume:

.m105.4m)39(

m105.2vol)

(

π πL

r πr L

10

)24.0()A10(V0

R r

)24.0()A30(V0

R r

energy is lost over the internal resistance

T

a ρ ρ T

ρ

dρ T

ndT T

n dT

0.8)K293()m10

5.3(  5  0 15   5   0 15

100.8:K77C

15 0

100.8:K573C

15 0

Trial and error shows that the right-hand side (rhs) above, for specific V values, equals

1333 V, when V 0.179V The current then is just

A

80.1]1)179.0(6.39[exp)A105.1(]16.39[

25.83: a) x L dx

A

R A

dx L x A

dx dR

A

L R

L

][exp]

1(]]

exp[

0

0 0

1 0

A V R

V I e

A

L L

x L

e V A

e I A

Le I x x

IR x

V x

E

L x L

x L



c)

)1()

1()

0()

1()

1

0 0

e V C

C e

V V

V C e

e V x

)1(

)(

)( 0  / 11

V x

.2

12

1

circuit short

R r

2 2 2

r

r r R